尽管在 Python 中进行了克隆，但列表元素发生了变化

Question

In the following Python code, I created a list a which contains a number of lists, and then created a copy of it, b, but unlike with lists that do not contain other lists, when b is changed, a also changes accordingly. Can someone explain how to prevent this from happening other than by copying each list separately?

a = [[None, None, None], [None, None, None], [None, None, None]]
b=a[:]
b[0][0] = 5
a
[[5, None, None], [None, None, None], [None, None, None]]

Answer 1

b is a shallow copy of a (check out the values id(a[0]) and id(b[0]) , they will be same even if id(a) != id(b) ). You need to take care of the elements inside the list as well (if they are mutable and modified, a shallow copy will reflect the changes in both the places). You can do something like this to create a deep copy:

import copy
b = copy.deepcopy(a)

Read up on shallow and deep copy operations in Python.

Answer 2

The slice creates a new list, but the contents of the list are not copied. So, a[:] gets you a new list, but the contents of that new list b are still references to the same lists that are referenced from a .

If you would try b[0]=[] , you'd be able to tell you can modify b without changing a , but b[0][0] just modifies the first element of the first list in b , which is the first element of the first list in a as well.